[LintCode] Valid Number

since1986 發布于2019-08-14 15:15 / 2574人閱讀

摘要：兩種情況有無，如上所示。如何判斷是否只有一個之后的長度小于等于。注意，為空只有在字符串最末位或者只有都是錯誤的。規則若字符串首位為，取其，然后將應用轉換為數組，逐位判斷字符，出現將置如已經置，再次出現則直接返回。

Problem

Validate if a given string is numeric.

Example

"0" => true

" 0.1 " => true

"abc" => false

"1 a" => false

"2e10" => true

Note

if !e
A can be "+/-" + int + "." + int

if e: AeB
A can be "+/-" + int + "." + int
B can be "+/-" + int

兩種情況：有無e，如上所示。
然后按有無e來拆分字符串，若拆分之后只有一部分，即用helper()函數判斷是否滿足構成數的規則；若有兩部分，則對前一部分按數A（可以有小數部分）分析，后一部分按構成數B分析（只能為整數）。
如何判斷是否只有一個e？split()之后的String[]長度小于等于2。
由于數A和數B的區別只存在于有沒有小數點"."，建立一個flag hasDot()，其他應用一樣的規則即可。注意，s為空、只有"e"、"e"在字符串最末位、或者只有"."都是錯誤的。

規則：若字符串首位為+/-，取其.substring(1)，然后將string應用toCharArray()轉換為char數組，逐位判斷字符，出現"."將hasDot()置true；如hasDot()已經置true，再次出現"."則直接返回false。然后判斷其余位是否為數字即可。

Solution

public class Solution {

public boolean isNumber(String s) {
    s = s.trim();
    if (s == null || (s.length() > 0 && s.charAt(s.length()-1) == "e") || s.equals(".")) return false;
    String[] strs = s.split("e");
    if (strs.length > 2) return false;
    boolean res = helper(strs[0], false);
    if (strs.length == 2) {
        res = res && helper(strs[1], true);
    }
    return res;
}
public boolean helper(String s, boolean hasDot) {
    if (s.length() > 0 && (s.charAt(0) == "+" || s.charAt(0) == "-"))
        s = s.substring(1);
        char[] ch = s.toCharArray();
        if (ch.length == 0) return false;
        for (int i =  0; i < ch.length; i++) {
            if (ch[i] == ".") {
                if (hasDot) return false;
                hasDot = true;
            }
            else if (!(ch[i] >= "0" && ch[i] <= "9")) return false;
        }
        return true;
}

}

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